Problem: $-3de + df - d - 3 = -7e + 2$ Solve for $d$.
Answer: Combine constant terms on the right. $-3de + df - d - {3} = -7e + {2}$ $-3de + df - d = -7e + {5}$ Notice that all the terms on the left-hand side of the equation have $d$ in them. $-3{d}e + 1{d}f - 1{d} = -7e + 5$ Factor out the $d$ ${d} \cdot \left( -3e + f - 1 \right) = -7e + 5$ Isolate the $d$ $d \cdot \left( -{3e + f - 1} \right) = -7e + 5$ $d = \dfrac{ -7e + 5 }{ -{3e + f - 1} }$ We can simplify this by multiplying the top and bottom by $-1$. $d= \dfrac{7e - 5}{3e - f + 1}$